TransWikia.com

Let $def*#1{mathbf{#1}}*{u},*{v},*{w}inmathbb{R}^2$ be noncollinear. Show that $*{x}= r*{u}+s*{v}+t*{w}$ where $r+s+t=1$.

Mathematics Asked on January 8, 2021

Suppose $mathbf{u},mathbf{v},mathbf{w}$ are noncollinear points $inmathbb{R}^2$. Let $mathbf{x} in mathbb{R}^2$. Show that we can write $mathbf{x}$ uniquely in the form $mathbf{x} = rmathbf{u}+smathbf{v}+tmathbf{w}$ where $r+s+t=1$.

Textbook (Shifrin’s Multivariable Mathematics) hint (paraphrased):

$mathbf{w}-mathbf{u} nparallel mathbf{v}-mathbf{u}$ so, $begin{bmatrix} mathbf{w}-mathbf{u}, mathbf{v}-mathbf{u}end{bmatrix}$ is an invertible $2times2$ matrix.


Attempt:

Let $mathbf{A} = begin{bmatrix} mathbf{w}-mathbf{u}, mathbf{v}-mathbf{u}end{bmatrix}, mathbf{c} = mathbf{A}^{-1} cdot mathbf{x}$.

$$
begin{align}
mathbf{x} &= mathbf{A}(mathbf{A}^{-1}mathbf{x}) \
&= mathbf{A}cdotmathbf{c} \
&= (mathbf{w}-mathbf{u})times c_1 + (mathbf{v}-mathbf{u})times c_2 \
&= c_1mathbf{w} + (-c_1-c_2)mathbf{u} + c_2mathbf{v}
end{align}
$$

Regrettably, $c_1 + (-c_1-c_2) + c_2 = 0 neq 1$.


Questions:

  1. Where did I go wrong?

2 Answers

Note that for any $x$ there is a unique $c$ such that $x-u = c_1 (w-u)+c_2(v-u)$.

In other words, $x = c_1 w + c_2 v + (1-c_1-c_2) u$.

Correct answer by copper.hat on January 8, 2021

You've found a representation of $mathbf x$ where the coefficients of $mathbf u, mathbf v, mathbf w$ add to $0$. That doesn't do everything you want, but it is a good first step.

Can you find a representation of $mathbf 0$ in which the coefficients of $mathbf u, mathbf v, mathbf w$ add to $1$? If so, then you could combine those, and get a representation of $mathbf x + mathbf 0$ in which the coefficients add to $0+1$.

Answered by Misha Lavrov on January 8, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP